I asked this question in this post but have not got a full answer. So I post it again on MO.

Consider the complex coefficient polynomial equation
\begin{eqnarray}
x^n-\left(a_1+\binom{n}{1}\right)x^{n-1}+\cdots+(-1)^k\left(a_k+\binom{n}{k}\right)x^{n-k}+\cdots+(-1)^{n-1}\left(a_{n-1}+\binom{n}{n-1}\right)x+(-1)^n=0
\end{eqnarray}
By Vieta Theorem, the product of its roots is 1. If we impose the condition that, among the $n$ roots, there exist $k$ roots (counted with multiplicity) whose product is 1, then $a_1, \cdots, a_{n-1}$ have to satisfy a polynomial equation $P(a_1, \cdots, a_{n-1})=0$, where $P\in\mathbb{C}[a_1, \cdots, a_{n-1}]$ has 0 as the constant term.

Question: Under what condition does $P$ has nonzero linear term?

Equivalently, I am asking for condition under which the affine hypersurface $V(P)\subset \mathbb{A}^{n-1}$ is smooth at the point $(0, 0, \cdots, 0)$.

The following are some easy examples I have worked out.

If $k=1$, then that means one of the roots is 1. Plugging $x=1$ to the original polynomial equation yields that $P$ can be
\begin{eqnarray}
\sum_{i=1}^{n-1} (-1)^ia_i
\end{eqnarray}
whose linear term is nonzero. For $k=2$, $n=3$ or $4$, $P$ also has nonzero linear term.

If $k=2$ and $n=5$, then the original polynomial equation can be factored as